Worked examples · Sheet 1

Classifying conics with the invariants

Theory used: Matrix associated with a conic Orthogonal invariants Degenerate conic Classification of non-degenerate conics Circles and rectangular hyperbolas

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Classify the conic

3 x^{2} + 2 x y + 3 y^{2} - 8 = 0.

A = 310130 00 - 8, A_{0} = (3113)

I_{3} = det A = - 8 \cdot (9 - 1) = - 64, I_{2} = det A_{0} = 8, I_{1} = tr A_{0} = 6

I_{3} \neq = 0, I_{2} > 0, I_{1} \cdot I_{3} = - 384 < 0

The conic is a real ellipse.

The coefficient of $x y$ is $2$ , so the matrix contains $a_{12} = 1$ ; there are no first-degree terms, therefore $a_{13} = a_{23} = 0$ and $a_{33} = - 8$ . Since the third row and the third column have a single nonzero entry, the determinant is computed by expanding along the third row and equals $- 8$ times the determinant of the block $A_{0}$ . Since $I_{3} \neq = 0$ , the conic is non-degenerate; since $I_{2} > 0$ , it is an ellipse, and since $I_{1}$ and $I_{3}$ have opposite signs, by the classification criterion the ellipse is real. It is not a circle, because $a_{12} = 1 \neq = 0$ .

Exercise 2

Classify the conic

x^{2} + 4 x y + y^{2} + 2 x - 2 y - 1 = 0.

A = 121 21 - 1 1 - 1 - 1

I_{3} = 1 \cdot (1 \cdot (- 1) - (- 1) (- 1)) - 2 \cdot (2 \cdot (- 1) - (- 1) \cdot 1) + 1 \cdot (2 \cdot (- 1) - 1 \cdot 1) = - 2 + 2 - 3 = - 3

I_{2} = 1221 = - 3, I_{1} = 2

The conic is a hyperbola, not rectangular.

The coefficients of $x y$ , $x$ and $y$ are $4$ , $2$ and $- 2$ , so the matrix contains their halves $a_{12} = 2$ , $a_{13} = 1$ , $a_{23} = - 1$ . The determinant of $A$ is computed by Laplace expansion along the first row and equals $- 3$ , so the conic is non-degenerate. Since $I_{2} = - 3 < 0$ , the conic is a hyperbola. A hyperbola is rectangular if and only if $I_{1} = 0$ ; here $I_{1} = 2$ , therefore the hyperbola is not rectangular.

Exercise 3

Classify the conic

x^{2} - 4 x y + 4 y^{2} + 2 x + 4 y = 0.

A = 1 - 2 1 - 2 42 120

I_{3} = 1 \cdot (0 - 4) + 2 \cdot (0 - 2) + 1 \cdot (- 4 - 4) = - 4 - 4 - 8 = - 16

I_{2} = 1 - 2 - 2 4 = 4 - 4 = 0

The conic is a parabola.

The matrix has $a_{12} = - 2$ , $a_{13} = 1$ and $a_{23} = 2$ . The expansion along the first row gives three terms: $1$ times the minor $4 \cdot 0 - 2 \cdot 2 = - 4$ , then $- (- 2) = 2$ times the minor $(- 2) \cdot 0 - 2 \cdot 1 = - 2$ , and finally $1$ times the minor $(- 2) \cdot 2 - 4 \cdot 1 = - 8$ . Hence $I_{3} = - 16 \neq = 0$ and the conic is non-degenerate. Since $I_{2} = 0$ , the conic is a parabola. The block $A_{0}$ has zero determinant because the second-degree part is the square $(x - 2 y)^{2}$ ; this is a quick check of the result.

Exercise 4

Classify the conic

x^{2} + 2 x y + 3 y^{2} + 2 x + 6 y + 7 = 0.

A = 111133137

I_{3} = 1 \cdot (21 - 9) - 1 \cdot (7 - 3) + 1 \cdot (3 - 3) = 12 - 4 + 0 = 8

I_{2} = 3 - 1 = 2 > 0, I_{1} = 4, I_{1} \cdot I_{3} = 32 > 0

{x + y + 1 = 0 x + 3 y + 3 = 0 ⟹ (0, - 1), f (0, - 1) = 3 - 6 + 7 = 4 = \frac{I _{3}}{I _{2}}

The conic is an imaginary ellipse: it has no real points.

The determinant of $A$ equals $8$ , so the conic is non-degenerate, and since $I_{2} = 2 > 0$ it is an ellipse. The product $I_{1} I_{3}$ is positive, therefore by the classification criterion the ellipse is imaginary. The result is checked with the canonical form: the center solves the system of the center and equals $(0, - 1)$ , and the value of the polynomial at the center is $4$ , equal to $\frac{I _{3}}{I _{2}}$ . The canonical form is therefore $λ_{1} X^{2} + λ_{2} Y^{2} + 4 = 0$ with $λ_{1}, λ_{2} > 0$ , since their product $I_{2}$ and their sum $I_{1}$ are positive. The left-hand side is at least $4$ for every $(X, Y)$ , so the equation has no real solutions.